Proving the limit of a multivariable function

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
7 replies · 3K views
SiriusAboutAstronomy
Messages
14
Reaction score
0

Homework Statement


Find the limit if it exists, or show that the limit does not exist.
lim (x,y)-> (1,0) (xy-y)/((x-1)^2+y^2)

Homework Equations


lim (x,y)-> (a,b) f(x,y)
0<((x-a)^2+(y-b)^2)^1/2<[itex]\delta[/itex]
abs(f(x,y)-L)<[itex]\epsilon[/itex]

The Attempt at a Solution


I tried to prove that it does not exist by analyzing the limit coming in from the x & y axes, and along lines y=x, yatta yatta. I kept getting 0, so I then tried to prove the limit exists and equals zero using the delta epsilon method. There I ran into problems, I have a total of 3 calculus books, each only has one example for the method and they are all the same example, which is also the same and only example that was covered in my class. lim (x,y) -> (0,0) (3yx^2)/(x^2+y^2). I am just looking for a starting point.

So far I have
0<((x-1)^2+(y)^2)^1/2<[itex]\delta[/itex]
abs((xy-y)/((x-1)^2+y^2))<[itex]\epsilon[/itex]
I know I need to manipulate it so that I can relate delta to be some multiple of epsilon, but don't know how.
 
Physics news on Phys.org
ConradYoung said:

Homework Statement


Find the limit if it exists, or show that the limit does not exist.
lim (x,y)-> (1,0) (xy-y)/((x-1)^2+y^2)

Homework Equations


lim (x,y)-> (a,b) f(x,y)
0<((x-a)^2+(y-b)^2)^1/2<[itex]\delta[/itex]
abs(f(x,y)-L)<[itex]\epsilon[/itex]

The Attempt at a Solution


I tried to prove that it does not exist by analyzing the limit coming in from the x & y axes, and along lines y=x, yatta yatta. I kept getting 0, so I then tried to prove the limit exists and equals zero using the delta epsilon method. There I ran into problems, I have a total of 3 calculus books, each only has one example for the method and they are all the same example, which is also the same and only example that was covered in my class. lim (x,y) -> (0,0) (3yx^2)/(x^2+y^2). I am just looking for a starting point.

So far I have
0<((x-1)^2+(y)^2)^1/2<[itex]\delta[/itex]
abs((xy-y)/((x-1)^2+y^2))<[itex]\epsilon[/itex]
I know I need to manipulate it so that I can relate delta to be some multiple of epsilon, but don't know how.
Try another path.

y = m(x-1) approaches the point (1, 0) along a line of arbitrary slope.
 
ConradYoung said:

Homework Statement


Find the limit if it exists, or show that the limit does not exist.
lim (x,y)-> (1,0) (xy-y)/((x-1)^2+y^2)

Homework Equations


lim (x,y)-> (a,b) f(x,y)
0<((x-a)^2+(y-b)^2)^1/2<[itex]\delta[/itex]
abs(f(x,y)-L)<[itex]\epsilon[/itex]

The Attempt at a Solution


I tried to prove that it does not exist by analyzing the limit coming in from the x & y axes, and along lines y=x, yatta yatta. I kept getting 0, so I then tried to prove the limit exists and equals zero using the delta epsilon method. There I ran into problems, I have a total of 3 calculus books, each only has one example for the method and they are all the same example, which is also the same and only example that was covered in my class. lim (x,y) -> (0,0) (3yx^2)/(x^2+y^2). I am just looking for a starting point.

So far I have
0<((x-1)^2+(y)^2)^1/2<[itex]\delta[/itex]
abs((xy-y)/((x-1)^2+y^2))<[itex]\epsilon[/itex]
I know I need to manipulate it so that I can relate delta to be some multiple of epsilon, but don't know how.

I'd first try a change of variables. Put u=x-1. So the limit is now (u,y)->(0,0). Denominator becomes u^2+y^2. What's the numerator? It might be easier to see that way.
 
Last edited:
Thank both of you.
 
Would either of you mind giving an example of using the delta epsilon method anyway?
 
Would either of you mind giving a different* example of using the delta epsilon method anyway?
 
ConradYoung said:
Would either of you mind giving a different* example of using the delta epsilon method anyway?

Ok, take f(x,y)=4xy/sqrt(x^2+y^2). (x,y)->(0,0). Change it to polar coordinates. You get |4r*cos(θ)*r*sin(θ)/r|=|4r*cos(θ)*sin(θ)|<=4r. So the limit is 0. And r is the "((x-a)^2+(y-b)^2)^1/2" in your definition. Pick ε>0. You want |f(x,y)-0|<ε, |f(x,y)-0|<4r, so if you pick δ=ε/4 it works.